In an alpha particle scattering experiment distance of closest approach for the $$\alpha$$ particle is $$4.5 \times 10^{-14} \mathrm{~m}$$. If target nucleus has atomic number 80 , then maximum velocity of $$\alpha$$-particle is __________ $$\times 10^5 \mathrm{~m} / \mathrm{s}$$ approximately.

($$\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}$$ unit, mass of $$\alpha$$ particle $$=6.72 \times 10^{-27} \mathrm{~kg}$$)